Question

For what value of \( k \) does the following pair of equations yield a unique solution for \( x \), such that the solution is positive?

\(x^2 - 3y^2 = 0 \)
\(x^2 - 6y^2 + k = 0 \)

• A. 2
• B. 0
• C. \( \sqrt{2} \)
• D. \( -\sqrt{2} \)

Hint:

Substitute from one equation into the other and solve for the condition that ensures a unique positive root.

Answer 1

The Correct Option is C

From the first equation: \( x^2 = 3y^2 \Rightarrow y^2 = \frac{x^2}{3} \)
Substitute into the second equation: \[ x^2 - 6\left( \frac{x^2}{3} \right) + k = 0 \Rightarrow x^2 - 2x^2 + k = 0 \Rightarrow -x^2 + k = 0 \Rightarrow x^2 = k \Rightarrow x = \sqrt{k} \] Now, for a **unique positive solution for \( x \)**, \( x = \sqrt{k} \Rightarrow k = x^2 \).
Let’s pick a specific positive value, say \( x = \sqrt{2} \Rightarrow k = 2 \), so: \[ x = \sqrt{2}, \quad k = 2 \] But we were asked the value of \( k \) such that **\( x = \sqrt{2} \)**. Hence, \[ {\sqrt{2}} \quad \text{is the correct value of } k \]